Question

Find the length of the spiraling polar curve

r=2e^(6theta)

From 0 to 2pi

Answer 1

Recall that arc length of a curve \(\mathcal{C}\) defined by a function \(\begin{cases}y=y(t)\\x=x(t)\end{cases}\) over an interval \(a\le t\le b\) is given by
\[\int_{\mathcal{C}}\mathrm{d}S=\int_{a}^{b}\sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2+\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2}\,\mathrm{d}t\]
To convert to polar coordinates, use the usual change of variables:
\[\begin{cases}x=r(t)\cos t\\y=r(t)\sin t\end{cases}~\implies~\begin{cases}
\dfrac{\mathrm{d}x}{\mathrm{d}t}=\dfrac{\mathrm{d}r}{\mathrm{d}t}\cos t-r\sin t\\[1ex]\dfrac{\mathrm{d}y}{\mathrm{d}t}=\dfrac{\mathrm{d}r}{\mathrm{d}t}\sin t+r\cos t\end{cases}\]which gives
\[\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2+\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2=r^2+\left(\frac{\mathrm{d}r}{\mathrm{d}t}\right)^2\]
So, the arc length of the curve \(\mathcal{C}\) given by the polar function defined by \(r(t)=2e^{6t}\) over \(0\le t\le2\pi\) is given by
\[\begin{align*}\int_{\mathcal{C}}\,\mathrm{d}S&=\int_0^{2\pi}\sqrt{\left(2e^{6t}\right)^2+\left(\frac{\mathrm{d}}{\mathrm{d}t}2e^{6t}\right)^2}\,\mathrm{d}t\\[1ex]
&=\int_0^{2\pi}\sqrt{148e^{12t}}\,\mathrm{d}t\\[1ex]
&=\sqrt{148}\int_0^{2\pi}e^{6t}\,\mathrm{d}t
\end{align*}\]